Question #66809
20cm3 of a gaseous hydrocarbon x were exploded with 120cm3 of oxygen. After the explosion the volume of the gases remaining was 90cm3 and this decreased to 50cm3 on treatment with aqueous potassium hydroxide. calculate the molecular formula of X
Expert's answer
If after the reaction remains O₂ and CO₂, that their volume has 90sm²
90-50=40sm³Progressed with alkali
50sm³(O₂) stay
120-50=70sm³(O₂)progressed
2CxHy+7O₂->4CO₂+H₂O The coefficients are the ratios of gas volumes
20(CxHy):70(O₂);40(CO₂)
The oxygen equation
2CxHy+7O₂->4CO₂+6H₂O
X=2;y=6
C₂H₆ - ethane
90-50=40sm³Progressed with alkali
50sm³(O₂) stay
120-50=70sm³(O₂)progressed
2CxHy+7O₂->4CO₂+H₂O The coefficients are the ratios of gas volumes
20(CxHy):70(O₂);40(CO₂)
The oxygen equation
2CxHy+7O₂->4CO₂+6H₂O
X=2;y=6
C₂H₆ - ethane
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